Unexpected behavior of operator "in" when checking if a list/tuple/etc. contains a value:
>>> 1 in [1] is True
False
>>> (1 in [1]) is True
True

Is this a bug? If not, please explain why first variant return False.

Check out this section of the documentation, notably this part:

"Note that comparisons, membership tests, and identity tests, all have the same precedence and have a left-to-right chaining feature"

Chaining lets you write stuff like this:

>>> x = 1
>>> 0 < x < 2
True

And since membership tests and identity tests are chained, the code you posted above essentially turns into:

(1 in [1]) and ([1] is True)

The former part of that expression is True but the latter is false.

Sorry, forgot the actual link. https://docs.python.org/3/reference/expressions.html#operator-precedence

Not a bug.  For an explanation, I just answered a very similar question on StackOverflow:

https://stackoverflow.com/questions/45180899/unexpected-result-from-in-operator/45180967#45180899

Thank you! It was my fault: I was not expecting `in` to be a comparison operator.

I am sure that some time ago I read that `in` is a comparison operator but I forgot it and I was thinking that (x in y) would be equivalent to (replaced with) the return value of y.__contains__(x).